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Proof step 10 from proof byte2bits-val (previous step)

# int2bits(b2i(by)) ≤ 8, b2i(by) < 0 ⊦ bits2uint(mklist(true, 8 - # int2bits(b2i(by))) + int2bits(b2i(by))) = b2i(by) + 256

Heuristic elimination applied rule insert elim lemma

And obtained the following new goals

Proof step 11:
8 = n + # int2bits(b2i(by)), # int2bits(b2i(by)) ≤ 8, b2i(by) < 0 ⊦ bits2uint(mklist(true, n) + int2bits(b2i(by))) = b2i(by) + 256

Rule argument

seq: n ≤ m ⊦ n0 = m - n ↔ m = n0 + n subst: [m, n, n0] → [8, # int2bits(b2i(by)), n] rotate: T spec: nat inst: name: elim-sub

Additional information

Used simplifier rules: n ≤ m ⊦ n0 = m - n ↔ m = n0 + n; (elim) Used terms: 8, # int2bits(b2i(by)), n